Proofs involving the addition of natural numbers
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This article contains mathematical proofs for some properties of addition of the natural numbers: the additive identity, commutativity, and associativity. These proofs are used in the article Addition of natural numbers.
Definitions
[edit]This article will use the Peano axioms for the definition of natural numbers. With these axioms, addition is defined from the constant 0 and the successor function S(a) by the two rules
A1: | a 0 = a |
A2: | a S(b) = S(a b) |
For the proof of commutativity, it is useful to give the name "1" to the successor of 0; that is,
- 1 = S(0).
For every natural number a, one has
S(a) | ||
= | S(a 0) | [by A1] |
= | a S(0) | [by A2] |
= | a 1 | [by Def. of 1] |
Proof of associativity
[edit]We prove associativity by first fixing natural numbers a and b and applying induction on the natural number c.
For the base case c = 0,
- (a b) 0 = a b = a (b 0)
Each equation follows by definition [A1]; the first with a b, the second with b.
Now, for the induction. We assume the induction hypothesis, namely we assume that for some natural number c,
- (a b) c = a (b c)
Then it follows,
(a b) S(c) | ||
= | S((a b) c) | [by A2] |
= | S(a (b c)) | [by the induction hypothesis] |
= | a S(b c) | [by A2] |
= | a (b S(c)) | [by A2] |
In other words, the induction hypothesis holds for S(c). Therefore, the induction on c is complete.
Proof of identity element
[edit]Definition [A1] states directly that 0 is a right identity. We prove that 0 is a left identity by induction on the natural number a.
For the base case a = 0, 0 0 = 0 by definition [A1]. Now we assume the induction hypothesis, that 0 a = a. Then
0 S(a) | ||
= | S(0 a) | [by A2] |
= | S(a) | [by the induction hypothesis] |
This completes the induction on a.
Proof of commutativity
[edit]We prove commutativity (a b = b a) by applying induction on the natural number b. First we prove the base cases b = 0 and b = S(0) = 1 (i.e. we prove that 0 and 1 commute with everything).
The base case b = 0 follows immediately from the identity element property (0 is an additive identity), which has been proved above: a 0 = a = 0 a.
Next we will prove the base case b = 1, that 1 commutes with everything, i.e. for all natural numbers a, we have a 1 = 1 a. We will prove this by induction on a (an induction proof within an induction proof). We have proved that 0 commutes with everything, so in particular, 0 commutes with 1: for a = 0, we have 0 1 = 1 0. Now, suppose a 1 = 1 a. Then
S(a) 1 | ||
= | S(a) S(0) | [by Def. of 1] |
= | S(S(a) 0) | [by A2] |
= | S((a 1) 0) | [as shown above] |
= | S(a 1) | [by A1] |
= | S(1 a) | [by the induction hypothesis] |
= | 1 S(a) | [by A2] |
This completes the induction on a, and so we have proved the base case b = 1. Now, suppose that for all natural numbers a, we have a b = b a. We must show that for all natural numbers a, we have a S(b) = S(b) a. We have
a S(b) | ||
= | a (b 1) | [as shown above] |
= | (a b) 1 | [by associativity] |
= | (b a) 1 | [by the induction hypothesis] |
= | b (a 1) | [by associativity] |
= | b (1 a) | [by the base case b = 1] |
= | (b 1) a | [by associativity] |
= | S(b) a | [as shown above] |
This completes the induction on b.
See also
[edit]References
[edit]- Edmund Landau, Foundations of Analysis, Chelsea Pub Co. ISBN 0-8218-2693-X.